Problem 8.5. Eigenfrequency of a water tower

Eigenfrequency, f [Hz]=

Follow steps of Example 53.

Step 1.  Assume rigid foundation. Approximate natural frequency of the cantilever with uniform and concentrated masses.

Show frequency, f1 Hide frequency, f1

The natural frequency of a cantilever with uniform mass is

Eq.(8-49).

$f_{m,1}^{}=\sqrt{0.13}\sqrt{\frac{{\mathrm{\pi }}^{2}EI}{4m{L}^4}}= \sqrt{0.13}\sqrt{\frac{{\mathrm{\pi }}^2\times 1\times {10}^9}{4\times 250/9.81\times {40}^4}}=2.22 \mathrm{Hz}$

The natural frequency of a cantilever with top concentrated mass is 

See Example 53, Eq.(8-15) and Table 11.

$f_{M,1}^{}==\frac{{\displaystyle 1}}{{\displaystyle 2\mathrm{\pi }}}\sqrt{\frac{{\displaystyle k}}{{\displaystyle M}}}=\frac{{\displaystyle 1}}{{\displaystyle 2\mathrm{\pi }}}\sqrt{\frac{{\displaystyle 3EI}}{{\displaystyle M{L}^3}}} = \frac{{\displaystyle 1}}{{\displaystyle 2\mathrm{\pi }}}\sqrt{\frac{{\displaystyle 3\times 1\times {10}^9}}{{\displaystyle 10000/9.81\times {40}^3}}}=1.08 \mathrm{Hz}$

where k is determined as the force-deflection ratio of the cantilever subjected to a concentrated top force.

Apply Dunkerley’s approximation to take into consideration both of the masses.

Table 30.

$\frac{1}{f_1^2}=\frac{{\displaystyle 1}}{{\displaystyle f_{m,1}^2}}+\frac{{\displaystyle 1}}{{\displaystyle f_{M,1}^2}}= \frac{{\displaystyle 1}}{{\displaystyle 2.{22}_{}^2}}+\frac{{\displaystyle 1}}{{\displaystyle 1.{08}_{}^2}} \to f_1^{}=0.970 \mathrm{Hz}$

Step 2.  Assume elastic foundation. Approximate natural frequency of the cantilever with uniform and concentrated masses.

Show frequency, f2 Hide frequency, f2

The natural frequency of a beam supported by a spring with uniform mass is derived in Problem 8.6:

$f_{m,2}^{}=\frac{\sqrt{3}}{2\mathrm{\pi }L}\sqrt{\frac{k}{mL}}=\frac{\sqrt{3}}{2\mathrm{\pi }\times 40} \sqrt{\frac{300\times {10}^6}{250/9.81\times 40}}=3.74 \mathrm{Hz}$

The natural frequency of the beam with top concentrated mass is 

Eq.(8-15)

$f_{M,2}^{}=\frac{{\displaystyle 1}}{{\displaystyle 2\mathrm{\pi }}}\sqrt{\frac{{\displaystyle k}}{{\displaystyle M{L}^2}}} = \frac{{\displaystyle 1}}{{\displaystyle 2\mathrm{\pi }}}\sqrt{\frac{{\displaystyle 300\times {10}^6}}{{\displaystyle 10000/9.81\times {40}^2}}}=2.16 \mathrm{Hz}$

Apply Dunkerley’s approximation to consider both masses.

Table 30.

$\frac{1}{f_2^2}=\frac{{\displaystyle 1}}{{\displaystyle f_{m,2}^2}}+\frac{{\displaystyle 1}}{{\displaystyle f_{M,2}^2}}= \frac{{\displaystyle 1}}{{\displaystyle 3.{74}_{}^2}}+\frac{{\displaystyle 1}}{{\displaystyle 2.{16}_{}^2}} \to f_1^{}=1.87 \mathrm{Hz}$

Step 3.  Apply Föppl’s approximation to take both supports into account.

Show frequency, f Hide frequency, f

Table 30.

$\frac{1}{f_{}^2}=\frac{{\displaystyle 1}}{{\displaystyle f_1^2}}+\frac{{\displaystyle 1}}{{\displaystyle f_2^2}}= \frac{{\displaystyle 1}}{{\displaystyle 0.{97}_{}^2}}+\frac{{\displaystyle 1}}{{\displaystyle 1.{87}_{}^2}} \to \mathit{f}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{861}\mathbf{ }\mathbf{Hz}$

Follow steps of Example 53.

First rigid foundation is assumed. The natural frequency of a cantilever with uniform mass is

Eq.(8-49).

$f_{m,1}^{}=\sqrt{0.13}\sqrt{\frac{{\mathrm{\pi }}^{2}EI}{4m{L}^4}}= \sqrt{0.13}\sqrt{\frac{{\mathrm{\pi }}^2\times 1\times {10}^9}{4\times 250/9.81\times {40}^4}}=2.22 \mathrm{Hz}$

The natural frequency of a cantilever with top concentrated mass is 

See Example 53, Eq.(8-15) and Table 11.

$f_{M,1}^{}==\frac{{\displaystyle 1}}{{\displaystyle 2\mathrm{\pi }}}\sqrt{\frac{{\displaystyle k}}{{\displaystyle M}}}=\frac{{\displaystyle 1}}{{\displaystyle 2\mathrm{\pi }}}\sqrt{\frac{{\displaystyle 3EI}}{{\displaystyle M{L}^3}}} = \frac{{\displaystyle 1}}{{\displaystyle 2\mathrm{\pi }}}\sqrt{\frac{{\displaystyle 3\times 1\times {10}^9}}{{\displaystyle 10000/9.81\times {40}^3}}}=1.08 \mathrm{Hz}$

where k is determined as the force-deflection ratio of the cantilever subjected to a concentrated top force.

Now Dunkerley’s approximation is applied to take into consideration both of the masses.

Table 30.

$\frac{1}{f_1^2}=\frac{{\displaystyle 1}}{{\displaystyle f_{m,1}^2}}+\frac{{\displaystyle 1}}{{\displaystyle f_{M,1}^2}}= \frac{{\displaystyle 1}}{{\displaystyle 2.{22}_{}^2}}+\frac{{\displaystyle 1}}{{\displaystyle 1.{08}_{}^2}} \to f_1^{}=0.970 \mathrm{Hz}$

Second elastic foundation is assumed. The natural frequency of a beam supported by a spring with uniform mass is is derived in Problem 8.6:

$f_{m,2}^{}=\frac{\sqrt{3}}{2\mathrm{\pi }L}\sqrt{\frac{k}{mL}}=\frac{\sqrt{3}}{2\mathrm{\pi }\times 40} \sqrt{\frac{300\times {10}^6}{250/9.81\times 40}}=3.74 \mathrm{Hz}$

The natural frequency of the beam with top concentrated mass is 

Eq.(8-15)

$f_{M,2}^{}=\frac{{\displaystyle 1}}{{\displaystyle 2\mathrm{\pi }}}\sqrt{\frac{{\displaystyle k}}{{\displaystyle M{L}^2}}} = \frac{{\displaystyle 1}}{{\displaystyle 2\mathrm{\pi }}}\sqrt{\frac{{\displaystyle 300\times {10}^6}}{{\displaystyle 10000/9.81\times {40}^2}}}=2.16 \mathrm{Hz}$

Dunkerley’s formula approximatites the effect of both masses.

Table 30.

$\frac{1}{f_2^2}=\frac{{\displaystyle 1}}{{\displaystyle f_{m,2}^2}}+\frac{{\displaystyle 1}}{{\displaystyle f_{M,2}^2}}= \frac{{\displaystyle 1}}{{\displaystyle 3.{74}_{}^2}}+\frac{{\displaystyle 1}}{{\displaystyle 2.{16}_{}^2}} \to f_1^{}=1.87 \mathrm{Hz}$

Finally Föppl’s approximation is applyied to take both supports into account.

Table 30.

$\frac{1}{f_{}^2}=\frac{{\displaystyle 1}}{{\displaystyle f_1^2}}+\frac{{\displaystyle 1}}{{\displaystyle f_2^2}}= \frac{{\displaystyle 1}}{{\displaystyle 0.{97}_{}^2}}+\frac{{\displaystyle 1}}{{\displaystyle 1.{87}_{}^2}} \to \mathit{f}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{861}\mathbf{ }\mathit{H}\mathit{z}$